TITLE: Rational normal curves and tangent lines
QUESTION [6 upvotes]: Let $C,\Gamma\subset\mathbb{P}^n$ be degree $n$ rational normal curves in $\mathbb{P}^n$, such that for any $p\in C$ the tangent line $T_pC$ of $C$ at $p$ is tangent to $\Gamma$ as well. This means that there exists a point $q\in\Gamma$ such that $T_q\Gamma = T_pC$.
Clearly, if $n = 2$ this means that $C$ and $\Gamma$ are two conics having the same dual conic i.e. $C^{*} = \Gamma^{*}$, and therefore $C = \Gamma$.
Is it true that this property (any tangent line of $C$ is tangent to $\Gamma$ as well) forces $C = \Gamma$ ?

REPLY [4 votes]: I want to suggest a simpler argument than Jason's, but I am not sure whether it works in positive characteristic or not.
Assume $n > 2$.
Consider the surface $S(C)$ swept by tangent lines to $C$. It is easy to see that the surface is not normal, its normalization is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$, its map to $\mathbb{P}^n$ is given by (an incomplete) linear system $\mathcal{O}(1,n-1)$, such that $C$ is the image of the diagonal in $\mathbb{P}^1 \times \mathbb{P}^1$ and is equal to the singular locus of the surface.
Now, assume that every tangent to $C$ is also tangent to $\Gamma$. Then $S(C) \subset S(\Gamma)$, and since both surfaces are irreducible, we have $S(C) = S(\Gamma)$. But then $C = Sing(S(C)) = Sing(S(\Gamma)) = \Gamma$.